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In physics, relativistic angular momentum refers to the mathematical formalisms and physical concepts that define angular momentum in special relativity (SR) and general relativity (GR). The relativistic quantity is subtly different from the three-dimensional quantity in classical mechanics.

Angular momentum is an important dynamical quantity derived from position and momentum. It is a measure of an object's rotational motion and resistance to changes in its rotation. Also, in the same way momentum conservation corresponds to translational symmetry, angular momentum conservation corresponds to rotational symmetry – the connection between symmetries and conservation laws is made by Noether's theorem. While these concepts were originally discovered in classical mechanics, they are also true and significant in special and general relativity. In terms of abstract algebra, the invariance of angular momentum, four-momentum, and other symmetries in spacetime, are described by the Lorentz group, or more generally the Poincaré group.

Physical quantities that remain separate in classical physics are naturally combined in SR and GR by enforcing the postulates of relativity. Most notably, the space and time coordinates combine into the four-position, and energy and momentum combine into the four-momentum. The components of these four-vectors depend on the frame of reference used, and change under Lorentz transformations to other inertial frames or accelerated frames.

Relativistic angular momentum is less obvious. The classical definition of angular momentum is the cross product of position x with momentum p to obtain a pseudovector x × p, or alternatively as the exterior product to obtain a second order antisymmetric tensor x ∧ p. What does this combine with, if anything? There is another vector quantity not often discussed – it is the time-varying moment of mass polar-vector (not the moment of inertia) related to the boost of the centre of mass of the system, and this combines with the classical angular momentum pseudovector to form an antisymmetric tensor of second order, in exactly the same way as the electric field polar-vector combines with the magnetic field pseudovector to form the electromagnetic field antisymmetric tensor. For rotating mass–energy distributions (such as gyroscopes, planets, stars, and black holes) instead of point-like particles, the angular momentum tensor is expressed in terms of the stress–energy tensor of the rotating object.

In special relativity alone, in the rest frame of a spinning object, there is an intrinsic angular momentum analogous to the "spin" in quantum mechanics and relativistic quantum mechanics, although for an extended body rather than a point particle. In relativistic quantum mechanics, elementary particles have spin and this is an additional contribution to the orbital angular momentum operator, yielding the total angular momentum tensor operator. In any case, the intrinsic "spin" addition to the orbital angular momentum of an object can be expressed in terms of the Pauli–Lubanski pseudovector.

Definitions

Orbital 3d angular momentum

For reference and background, two closely related forms of angular momentum are given.

In classical mechanics, the orbital angular momentum of a particle with instantaneous three-dimensional position vector x = (x, y, z) and momentum vector p = (p_x, p_y, p_z), is defined as the axial vector $${\displaystyle \mathbf {L} =\mathbf {x} \times \mathbf {p} }$$ which has three components, that are systematically given by cyclic permutations of Cartesian directions (e.g. change x to y, y to z, z to x, repeat) $${\displaystyle {\begin{aligned}L_{x}&=yp_{z}-zp_{y}\,,\\L_{y}&=zp_{x}-xp_{z}\,,\\L_{z}&=xp_{y}-yp_{x}\,.\end{aligned}}}$$

A related definition is to conceive orbital angular momentum as a plane element. This can be achieved by replacing the cross product by the exterior product in the language of exterior algebra, and angular momentum becomes a contravariant second order antisymmetric tensor $${\displaystyle \mathbf {L} =\mathbf {x} \wedge \mathbf {p} }$$

or writing x = (x₁, x₂, x₃) = (x, y, z) and momentum vector p = (p₁, p₂, p₃) = (p_x, p_y, p_z), the components can be compactly abbreviated in tensor index notation $${\displaystyle L^{ij}=x^{i}p^{j}-x^{j}p^{i}}$$ where the indices i and j take the values 1, 2, 3. On the other hand, the components can be systematically displayed fully in a 3 × 3 antisymmetric matrix $${\displaystyle {\begin{aligned}\mathbf {L} &={\begin{pmatrix}L^{11}&L^{12}&L^{13}\\L^{21}&L^{22}&L^{23}\\L^{31}&L^{32}&L^{33}\\\end{pmatrix}}={\begin{pmatrix}0&L_{xy}&L_{xz}\\L_{yx}&0&L_{yz}\\L_{zx}&L_{zy}&0\end{pmatrix}}={\begin{pmatrix}0&L_{xy}&-L_{zx}\\-L_{xy}&0&L_{yz}\\L_{zx}&-L_{yz}&0\end{pmatrix}}\\&={\begin{pmatrix}0&xp_{y}-yp_{x}&-(zp_{x}-xp_{z})\\-(xp_{y}-yp_{x})&0&yp_{z}-zp_{y}\\zp_{x}-xp_{z}&-(yp_{z}-zp_{y})&0\end{pmatrix}}\end{aligned}}}$$

This quantity is additive, and for an isolated system, the total angular momentum of a system is conserved.

Dynamic mass moment

In classical mechanics, the three-dimensional quantity for a particle of mass m moving with velocity u $${\displaystyle \mathbf {N} =m\left(\mathbf {x} -t\mathbf {u} \right)=m\mathbf {x} -t\mathbf {p} }$$ has the dimensions of mass moment – length multiplied by mass. It is equal to the mass of the particle or system of particles multiplied by the distance from the space origin to the centre of mass (COM) at the time origin (t = 0), as measured in the lab frame. There is no universal symbol, nor even a universal name, for this quantity. Different authors may denote it by other symbols if any (for example μ), may designate other names, and may define N to be the negative of what is used here. The above form has the advantage that it resembles the familiar Galilean transformation for position, which in turn is the non-relativistic boost transformation between inertial frames.

This vector is also additive: for a system of particles, the vector sum is the resultant $${\displaystyle \sum _{n}\mathbf {N} _{n}=\sum _{n}m_{n}\left(\mathbf {x} _{n}-t\mathbf {u} _{n}\right)=\left(\mathbf {x} _{\mathrm {COM} }\sum _{n}m_{n}-t\sum _{n}m_{n}\mathbf {u} _{n}\right)=M_{\text{tot}}(\mathbf {x} _{\mathrm {COM} }-\mathbf {u} _{\mathrm {COM} }t)}$$ where the system's centre of mass position and velocity and total mass are respectively $${\displaystyle {\begin{aligned}\mathbf {x} _{\mathrm {COM} }&={\frac {\sum _{n}m_{n}\mathbf {x} _{n}}{\sum _{n}m_{n}}},\\\mathbf {u} _{\mathrm {COM} }&={\frac {\sum _{n}m_{n}\mathbf {u} _{n}}{\sum _{n}m_{n}}},\\M_{\text{tot}}&=\sum _{n}m_{n}.\end{aligned}}}$$

For an isolated system, N is conserved in time, which can be seen by differentiating with respect to time. The angular momentum L is a pseudovector, but N is an "ordinary" (polar) vector, and is therefore invariant under inversion.

The resultant N_tot for a multiparticle system has the physical visualization that, whatever the complicated motion of all the particles are, they move in such a way that the system's COM moves in a straight line. This does not necessarily mean all particles "follow" the COM, nor that all particles all move in almost the same direction simultaneously, only that the collective motion of the particles is constrained in relation to the centre of mass.

In special relativity, if the particle moves with velocity u relative to the lab frame, then $${\displaystyle {\begin{aligned}E&=\gamma (\mathbf {u} )m_{0}c^{2},&\mathbf {p} &=\gamma (\mathbf {u} )m_{0}\mathbf {u} \end{aligned}}}$$ where $${\displaystyle \gamma (\mathbf {u} )={\frac {1}{\sqrt {1-{\frac {\mathbf {u} \cdot \mathbf {u} }{c^{2}}}}}}}$$ is the Lorentz factor and m is the mass (i.e. the rest mass) of the particle. The corresponding relativistic mass moment in terms of m, u, p, E, in the same lab frame is $${\displaystyle \mathbf {N} ={\frac {E}{c^{2}}}\mathbf {x} -\mathbf {p} t=m\gamma (\mathbf {u} )(\mathbf {x} -\mathbf {u} t).}$$

The Cartesian components are $${\displaystyle {\begin{aligned}N_{x}=mx-p_{x}t&={\frac {E}{c^{2}}}x-p_{x}t=m\gamma (u)(x-u_{x}t)\\N_{y}=my-p_{y}t&={\frac {E}{c^{2}}}y-p_{y}t=m\gamma (u)(y-u_{y}t)\\N_{z}=mz-p_{z}t&={\frac {E}{c^{2}}}z-p_{z}t=m\gamma (u)(z-u_{z}t)\end{aligned}}}$$

Special relativity

Coordinate transformations for a boost in the x direction

Consider a coordinate frame F′ which moves with velocity v = (v, 0, 0) relative to another frame F, along the direction of the coincident xx′ axes. The origins of the two coordinate frames coincide at times t = t′ = 0. The mass–energy E = mc and momentum components p = (p_x, p_y, p_z) of an object, as well as position coordinates x = (x, y, z) and time t in frame F are transformed to E′ = m′c, p′ = (p_x′, p_y′, p_z′), x′ = (x′, y′, z′), and t′ in F′ according to the Lorentz transformations $${\displaystyle {\begin{aligned}t'&=\gamma (v)\left(t-{\frac {vx}{c^{2}}}\right)\,,\quad &E'&=\gamma (v)\left(E-vp_{x}\right)\\x'&=\gamma (v)(x-vt)\,,\quad &p_{x}'&=\gamma (v)\left(p_{x}-{\frac {vE}{c^{2}}}\right)\\y'&=y\,,\quad &p_{y}'&=p_{y}\\z'&=z\,,\quad &p_{z}'&=p_{z}\\\end{aligned}}}$$

The Lorentz factor here applies to the velocity v, the relative velocity between the frames. This is not necessarily the same as the velocity u of an object.

For the orbital 3-angular momentum L as a pseudovector, we have $${\displaystyle {\begin{aligned}L_{x}'&=y'p_{z}'-z'p_{y}'=L_{x}\\L_{y}'&=z'p_{x}'-x'p_{z}'=\gamma (v)(L_{y}-vN_{z})\\L_{z}'&=x'p_{y}'-y'p_{x}'=\gamma (v)(L_{z}+vN_{y})\\\end{aligned}}}$$

For the x-component $${\displaystyle L_{x}'=y'p_{z}'-z'p_{y}'=yp_{z}-zp_{y}=L_{x}}$$ the y-component $${\displaystyle {\begin{aligned}L_{y}'&=z'p_{x}'-x'p_{z}'\\&=z\gamma \left(p_{x}-{\frac {vE}{c^{2}}}\right)-\gamma \left(x-vt\right)p_{z}\\&=\gamma \left\\&=\gamma \left\\&=\gamma \left(L_{y}-vN_{z}\right)\end{aligned}}}$$ and z-component $${\displaystyle {\begin{aligned}L_{z}'&=x'p_{y}'-y'p_{x}'\\&=\gamma \left(x-vt\right)p_{y}-y\gamma \left(p_{x}-{\frac {vE}{c^{2}}}\right)\\&=\gamma \left\\&=\gamma \left\\&=\gamma \left(L_{z}+vN_{y}\right)\end{aligned}}}$$

In the second terms of L_y′ and L_z′, the y and z components of the cross product v × N can be inferred by recognizing cyclic permutations of v_x = v and v_y = v_z = 0 with the components of N, $${\displaystyle {\begin{aligned}-vN_{z}&=v_{z}N_{x}-v_{x}N_{z}=\left(\mathbf {v} \times \mathbf {N} \right)_{y}\\vN_{y}&=v_{x}N_{y}-v_{y}N_{x}=\left(\mathbf {v} \times \mathbf {N} \right)_{z}\\\end{aligned}}}$$

Now, L_x is parallel to the relative velocity v, and the other components L_y and L_z are perpendicular to v. The parallel–perpendicular correspondence can be facilitated by splitting the entire 3-angular momentum pseudovector into components parallel (∥) and perpendicular (⊥) to v, in each frame, $${\displaystyle \mathbf {L} =\mathbf {L} _{\parallel }+\mathbf {L} _{\perp }\,,\quad \mathbf {L} '=\mathbf {L} _{\parallel }'+\mathbf {L} _{\perp }'\,.}$$

Then the component equations can be collected into the pseudovector equations $${\displaystyle {\begin{aligned}\mathbf {L} _{\parallel }'&=\mathbf {L} _{\parallel }\\\mathbf {L} _{\perp }'&=\gamma (\mathbf {v} )\left(\mathbf {L} _{\perp }+\mathbf {v} \times \mathbf {N} \right)\\\end{aligned}}}$$

Therefore, the components of angular momentum along the direction of motion do not change, while the components perpendicular do change. By contrast to the transformations of space and time, time and the spatial coordinates change along the direction of motion, while those perpendicular do not.

These transformations are true for all v, not just for motion along the xx′ axes.

Considering L as a tensor, we get a similar result $${\displaystyle \mathbf {L} _{\perp }'=\gamma (\mathbf {v} )\left(\mathbf {L} _{\perp }+\mathbf {v} \wedge \mathbf {N} \right)}$$ where $${\displaystyle {\begin{aligned}v_{z}N_{x}-v_{x}N_{z}&=\left(\mathbf {v} \wedge \mathbf {N} \right)_{zx}\\v_{x}N_{y}-v_{y}N_{x}&=\left(\mathbf {v} \wedge \mathbf {N} \right)_{xy}\\\end{aligned}}}$$

The boost of the dynamic mass moment along the x direction is $${\displaystyle {\begin{aligned}N_{x}'&=m'x'-p_{x}'t'=N_{x}\\N_{y}'&=m'y'-p_{y}'t'=\gamma (v)\left(N_{y}+{\frac {vL_{z}}{c^{2}}}\right)\\N_{z}'&=m'z'-p_{z}'t'=\gamma (v)\left(N_{z}-{\frac {vL_{y}}{c^{2}}}\right)\\\end{aligned}}}$$

For the x-component $${\displaystyle {\begin{aligned}N_{x}'&={\frac {E'}{c^{2}}}x'-t'p_{x}'\\&={\frac {\gamma }{c^{2}}}(E-vp_{x})\gamma (x-vt)-\gamma \left(t-{\frac {xv}{c^{2}}}\right)\gamma \left(p_{x}-{\frac {vE}{c^{2}}}\right)\\&=\gamma ^{2}\left\\&=\gamma ^{2}\left\\&=\gamma ^{2}\left\\&=\gamma ^{2}\left\\&=\gamma ^{2}\leftN_{x}\\&=\gamma ^{2}{\frac {1}{\gamma ^{2}}}N_{x}\end{aligned}}}$$ the y-component $${\displaystyle {\begin{aligned}N_{y}'&={\frac {E'}{c^{2}}}y'-t'p_{y}'\\&={\frac {1}{c^{2}}}\gamma (E-vp_{x})y-\gamma \left(t-{\frac {xv}{c^{2}}}\right)p_{y}\\&=\gamma \left\\&=\gamma \left\\&=\gamma \left\\&=\gamma \left(N_{y}+{\frac {v}{c^{2}}}L_{z}\right)\end{aligned}}}$$ and z-component $${\displaystyle {\begin{aligned}N_{z}'&={\frac {E'}{c^{2}}}z'-t'p_{z}'\\&={\frac {1}{c^{2}}}\gamma (E-vp_{x})z-\gamma \left(t-{\frac {xv}{c^{2}}}\right)p_{z}\\&=\gamma \left\\&=\gamma \left\\&=\gamma \left\\&=\gamma \left(N_{z}-{\frac {v}{c^{2}}}L_{y}\right)\end{aligned}}}$$

Collecting parallel and perpendicular components as before $${\displaystyle {\begin{aligned}\mathbf {N} _{\parallel }'&=\mathbf {N} _{\parallel }\\\mathbf {N} _{\perp }'&=\gamma (\mathbf {v} )\left(\mathbf {N} _{\perp }-{\frac {1}{c^{2}}}\mathbf {v} \times \mathbf {L} \right)\\\end{aligned}}}$$

Again, the components parallel to the direction of relative motion do not change, those perpendicular do change.

Vector transformations for a boost in any direction

So far these are only the parallel and perpendicular decompositions of the vectors. The transformations on the full vectors can be constructed from them as follows (throughout here L is a pseudovector for concreteness and compatibility with vector algebra).

Introduce a unit vector in the direction of v, given by n = v/v. The parallel components are given by the vector projection of L or N into n $${\displaystyle \mathbf {L} _{\parallel }=(\mathbf {L} \cdot \mathbf {n} )\mathbf {n} \,,\quad \mathbf {N} _{\parallel }=(\mathbf {N} \cdot \mathbf {n} )\mathbf {n} }$$ while the perpendicular component by vector rejection of L or N from n $${\displaystyle \mathbf {L} _{\perp }=\mathbf {L} -(\mathbf {L} \cdot \mathbf {n} )\mathbf {n} \,,\quad \mathbf {N} _{\perp }=\mathbf {N} -(\mathbf {N} \cdot \mathbf {n} )\mathbf {n} }$$ and the transformations are $${\displaystyle {\begin{aligned}\mathbf {L} '&=\gamma (\mathbf {v} )(\mathbf {L} +v\mathbf {n} \times \mathbf {N} )-(\gamma (\mathbf {v} )-1)(\mathbf {L} \cdot \mathbf {n} )\mathbf {n} \\\mathbf {N} '&=\gamma (\mathbf {v} )\left(\mathbf {N} -{\frac {v}{c^{2}}}\mathbf {n} \times \mathbf {L} \right)-(\gamma (\mathbf {v} )-1)(\mathbf {N} \cdot \mathbf {n} )\mathbf {n} \\\end{aligned}}}$$ or reinstating v = vn, $${\displaystyle {\begin{aligned}\mathbf {L} '&=\gamma (\mathbf {v} )(\mathbf {L} +\mathbf {v} \times \mathbf {N} )-(\gamma (\mathbf {v} )-1){\frac {(\mathbf {L} \cdot \mathbf {v} )\mathbf {v} }{v^{2}}}\\\mathbf {N} '&=\gamma (\mathbf {v} )\left(\mathbf {N} -{\frac {1}{c^{2}}}\mathbf {v} \times \mathbf {L} \right)-(\gamma (\mathbf {v} )-1){\frac {(\mathbf {N} \cdot \mathbf {v} )\mathbf {v} }{v^{2}}}\\\end{aligned}}}$$

These are very similar to the Lorentz transformations of the electric field E and magnetic field B, see Classical electromagnetism and special relativity.

Alternatively, starting from the vector Lorentz transformations of time, space, energy, and momentum, for a boost with velocity v, $${\displaystyle {\begin{aligned}t'&=\gamma (\mathbf {v} )\left(t-{\frac {\mathbf {v} \cdot \mathbf {r} }{c^{2}}}\right)\,,\\\mathbf {r} '&=\mathbf {r} +{\frac {\gamma (\mathbf {v} )-1}{v^{2}}}(\mathbf {r} \cdot \mathbf {v} )\mathbf {v} -\gamma (\mathbf {v} )t\mathbf {v} \,,\\\mathbf {p} '&=\mathbf {p} +{\frac {\gamma (\mathbf {v} )-1}{v^{2}}}(\mathbf {p} \cdot \mathbf {v} )\mathbf {v} -\gamma (\mathbf {v} ){\frac {E}{c^{2}}}\mathbf {v} \,,\\E'&=\gamma (\mathbf {v} )\left(E-\mathbf {v} \cdot \mathbf {p} \right)\,,\\\end{aligned}}}$$ inserting these into the definitions $${\displaystyle {\begin{aligned}\mathbf {L} '&=\mathbf {r} '\times \mathbf {p} '\,,&\mathbf {N} '&={\frac {E'}{c^{2}}}\mathbf {r} '-t'\mathbf {p} '\end{aligned}}}$$ gives the transformations.

The orbital angular momentum in each frame are $${\displaystyle \mathbf {L} '=\mathbf {r} '\times \mathbf {p} '\,,\quad \mathbf {L} =\mathbf {r} \times \mathbf {p} }$$ so taking the cross product of the transformations $${\displaystyle {\begin{aligned}\mathbf {L} '&=\left\times \left\\&=\times \mathbf {p} +(\gamma -1)(\mathbf {p} \cdot \mathbf {n} )\times \mathbf {n} -\gamma {\frac {E}{c^{2}}}v\times \mathbf {n} \\&=\mathbf {r} \times \mathbf {p} +(\gamma -1)(\mathbf {r} \cdot \mathbf {n} )\mathbf {n} \times \mathbf {p} -\gamma tv\mathbf {n} \times \mathbf {p} +(\gamma -1)(\mathbf {p} \cdot \mathbf {n} )\mathbf {r} \times \mathbf {n} -\gamma {\frac {E}{c^{2}}}v\mathbf {r} \times \mathbf {n} \\&=\mathbf {L} +\left\mathbf {v} \times \mathbf {p} +\left\mathbf {r} \times \mathbf {v} \\&=\mathbf {L} +\mathbf {v} \times \left\\&=\mathbf {L} +\mathbf {v} \times \left\end{aligned}}}$$

Using the triple product rule $${\displaystyle {\begin{aligned}\mathbf {a} \times (\mathbf {b} \times \mathbf {c} )&=\mathbf {b} (\mathbf {a} \cdot \mathbf {c} )-\mathbf {c} (\mathbf {a} \cdot \mathbf {b} )\\(\mathbf {a} \times \mathbf {b} )\times \mathbf {c} &=(\mathbf {c} \cdot \mathbf {a} )\mathbf {b} -(\mathbf {c} \cdot \mathbf {b} )\mathbf {a} \\\end{aligned}}}$$ gives $${\displaystyle {\begin{aligned}(\mathbf {r} \times \mathbf {p} )\times \mathbf {v} &=(\mathbf {v} \cdot \mathbf {r} )\mathbf {p} -(\mathbf {v} \cdot \mathbf {p} )\mathbf {r} \\(\mathbf {v} \cdot \mathbf {r} )\mathbf {p} -(\mathbf {v} \cdot \mathbf {p} )\mathbf {r} &=\mathbf {L} \times \mathbf {v} \\\end{aligned}}}$$ and along with the definition of N we have $${\displaystyle \mathbf {L} '=\mathbf {L} +\mathbf {v} \times \left}$$

Reinstating the unit vector n, $${\displaystyle \mathbf {L} '=\mathbf {L} +\mathbf {n} \times \left}$$

Since in the transformation there is a cross product on the left with n, $${\displaystyle \mathbf {n} \times (\mathbf {L} \times \mathbf {n} )=\mathbf {L} (\mathbf {n} \cdot \mathbf {n} )-\mathbf {n} (\mathbf {n} \cdot \mathbf {L} )=\mathbf {L} -\mathbf {n} (\mathbf {n} \cdot \mathbf {L} )}$$ then $${\displaystyle \mathbf {L} '=\mathbf {L} +(\gamma -1)(\mathbf {L} -\mathbf {n} (\mathbf {n} \cdot \mathbf {L} ))+\beta \gamma c\mathbf {n} \times \mathbf {N} =\gamma (\mathbf {L} +v\mathbf {n} \times \mathbf {N} )-(\gamma -1)\mathbf {n} (\mathbf {n} \cdot \mathbf {L} )}$$

4d angular momentum as a bivector

In relativistic mechanics, the COM boost and orbital 3-space angular momentum of a rotating object are combined into a four-dimensional bivector in terms of the four-position X and the four-momentum P of the object $${\displaystyle \mathbf {M} =\mathbf {X} \wedge \mathbf {P} }$$

In components $${\displaystyle M^{\alpha \beta }=X^{\alpha }P^{\beta }-X^{\beta }P^{\alpha }}$$ which are six independent quantities altogether. Since the components of X and P are frame-dependent, so is M. Three components $${\displaystyle M^{ij}=x^{i}p^{j}-x^{j}p^{i}=L^{ij}}$$ are those of the familiar classical 3-space orbital angular momentum, and the other three $${\displaystyle M^{0i}=x^{0}p^{i}-x^{i}p^{0}=c\,\left(tp^{i}-x^{i}{\frac {E}{c^{2}}}\right)=-cN^{i}}$$ are the relativistic mass moment, multiplied by −c. The tensor is antisymmetric; $${\displaystyle M^{\alpha \beta }=-M^{\beta \alpha }}$$

The components of the tensor can be systematically displayed as a matrix $${\displaystyle {\begin{aligned}\mathbf {M} &={\begin{pmatrix}M^{00}&M^{01}&M^{02}&M^{03}\\M^{10}&M^{11}&M^{12}&M^{13}\\M^{20}&M^{21}&M^{22}&M^{23}\\M^{30}&M^{31}&M^{32}&M^{33}\end{pmatrix}}\\&=\left({\begin{array}{c|ccc}0&-N^{1}c&-N^{2}c&-N^{3}c\\\hline N^{1}c&0&L^{12}&-L^{31}\\N^{2}c&-L^{12}&0&L^{23}\\N^{3}c&L^{31}&-L^{23}&0\end{array}}\right)\\&=\left({\begin{array}{c|c}0&-\mathbf {N} c\\\hline \mathbf {N} ^{\mathrm {T} }c&\mathbf {x} \wedge \mathbf {p} \\\end{array}}\right)\end{aligned}}}$$ in which the last array is a block matrix formed by treating N as a row vector which matrix transposes to the column vector N, and x ∧ p as a 3 × 3 antisymmetric matrix. The lines are merely inserted to show where the blocks are.

Again, this tensor is additive: the total angular momentum of a system is the sum of the angular momentum tensors for each constituent of the system: $${\displaystyle \mathbf {M} _{\text{tot}}=\sum _{n}\mathbf {M} _{n}=\sum _{n}\mathbf {X} _{n}\wedge \mathbf {P} _{n}\,.}$$

Each of the six components forms a conserved quantity when aggregated with the corresponding components for other objects and fields.

The angular momentum tensor M is indeed a tensor, the components change according to a Lorentz transformation matrix Λ, as illustrated in the usual way by tensor index notation $${\displaystyle {\begin{aligned}{M'}^{\alpha \beta }&={X'}^{\alpha }{P'}^{\beta }-{X'}^{\beta }{P'}^{\alpha }\\&={\Lambda ^{\alpha }}_{\gamma }X^{\gamma }{\Lambda ^{\beta }}_{\delta }P^{\delta }-{\Lambda ^{\beta }}_{\delta }X^{\delta }{\Lambda ^{\alpha }}_{\gamma }P^{\gamma }\\&={\Lambda ^{\alpha }}_{\gamma }{\Lambda ^{\beta }}_{\delta }\left(X^{\gamma }P^{\delta }-X^{\delta }P^{\gamma }\right)\\&={\Lambda ^{\alpha }}_{\gamma }{\Lambda ^{\beta }}_{\delta }M^{\gamma \delta }\\\end{aligned}},}$$ where, for a boost (without rotations) with normalized velocity β = v/c, the Lorentz transformation matrix elements are $${\displaystyle {\begin{aligned}{\Lambda ^{0}}_{0}&=\gamma \\{\Lambda ^{i}}_{0}&={\Lambda ^{0}}_{i}=-\gamma \beta ^{i}\\{\Lambda ^{i}}_{j}&={\delta ^{i}}_{j}+{\frac {\gamma -1}{\beta ^{2}}}\beta ^{i}\beta _{j}\end{aligned}}}$$ and the covariant β_i and contravariant β components of β are the same since these are just parameters.

In other words, one can Lorentz-transform the four position and four momentum separately, and then antisymmetrize those newly found components to obtain the angular momentum tensor in the new frame.

The transformation of boost components are

$${\displaystyle {\begin{aligned}M'^{k0}&={\Lambda ^{k}}_{\mu }{\Lambda ^{0}}_{\nu }M^{\mu \nu }\\&={\Lambda ^{k}}_{0}{\Lambda ^{0}}_{0}M^{00}+{\Lambda ^{k}}_{i}{\Lambda ^{0}}_{0}M^{i0}+{\Lambda ^{k}}_{0}{\Lambda ^{0}}_{j}M^{0j}+{\Lambda ^{k}}_{i}{\Lambda ^{0}}_{j}M^{ij}\\&=\left({\Lambda ^{k}}_{i}{\Lambda ^{0}}_{0}-{\Lambda ^{k}}_{0}{\Lambda ^{0}}_{i}\right)M^{i0}+{\Lambda ^{k}}_{i}{\Lambda ^{0}}_{j}M^{ij}\\\end{aligned}}}$$ as for the orbital angular momentum $${\displaystyle {\begin{aligned}{M'}^{k\ell }&={\Lambda ^{k}}_{\mu }{\Lambda ^{\ell }}_{\nu }M^{\mu \nu }\\&={\Lambda ^{k}}_{0}{\Lambda ^{\ell }}_{0}M^{00}+{\Lambda ^{k}}_{i}{\Lambda ^{\ell }}_{0}M^{i0}+{\Lambda ^{k}}_{0}{\Lambda ^{\ell }}_{j}M^{0j}+{\Lambda ^{k}}_{i}{\Lambda ^{\ell }}_{j}M^{ij}\\&=\left({\Lambda ^{k}}_{i}{\Lambda ^{\ell }}_{0}-{\Lambda ^{k}}_{0}{\Lambda ^{\ell }}_{i}\right)M^{i0}+{\Lambda ^{k}}_{i}{\Lambda ^{\ell }}_{j}M^{ij}\end{aligned}}}$$

The expressions in the Lorentz transformation entries are $${\displaystyle {\begin{aligned}{\Lambda ^{k}}_{i}{\Lambda ^{\ell }}_{0}-{\Lambda ^{k}}_{0}{\Lambda ^{\ell }}_{i}&=\left\left(-\gamma \beta ^{\ell }\right)-\left(-\gamma \beta ^{k}\right)\left\\&=\gamma \left\\{\Lambda ^{k}}_{i}{\Lambda ^{0}}_{0}-{\Lambda ^{k}}_{0}{\Lambda ^{0}}_{i}&=\left\gamma -(-\gamma \beta ^{k})(-\gamma \beta ^{i})\\&=\gamma \left\\&=\gamma \left\\&=\gamma \left\\&=\gamma \left\\&=\gamma {\delta ^{k}}_{i}-\left\beta ^{k}\beta _{i}\\{\Lambda ^{k}}_{i}{\Lambda ^{\ell }}_{j}&=\left\left\\&={\delta ^{k}}_{i}{\delta ^{\ell }}_{j}+{\frac {\gamma -1}{\beta ^{2}}}{\delta ^{k}}_{i}\beta ^{\ell }\beta _{j}+{\frac {\gamma -1}{\beta ^{2}}}\beta ^{k}\beta _{i}{\delta ^{\ell }}_{j}+{\frac {\gamma -1}{\beta ^{2}}}{\frac {\gamma -1}{\beta ^{2}}}\beta ^{\ell }\beta _{j}\beta ^{k}\beta _{i}\end{aligned}}}$$ gives $${\displaystyle {\begin{aligned}cN'^{k}&=\left({\Lambda ^{k}}_{i}{\Lambda ^{0}}_{0}-{\Lambda ^{k}}_{0}{\Lambda ^{0}}_{i}\right)cN^{i}+{\Lambda ^{k}}_{i}{\Lambda ^{0}}_{j}\varepsilon ^{ijn}L_{n}\\&=\leftcN^{i}+-\gamma \beta ^{j}\left\varepsilon ^{ijn}L_{n}\\&=\gamma cN^{k}-\left({\frac {\gamma -1}{\beta ^{2}}}\right)\beta ^{k}\left(\beta _{i}cN^{i}\right)-\gamma \beta ^{j}{\delta ^{k}}_{i}\varepsilon ^{ijn}L_{n}-\gamma {\frac {\gamma -1}{\beta ^{2}}}\beta ^{j}\beta ^{k}\beta _{i}\varepsilon ^{ijn}L_{n}\\&=\gamma cN^{k}-\left({\frac {\gamma -1}{\beta ^{2}}}\right)\beta ^{k}\left(\beta _{i}cN^{i}\right)-\gamma \beta ^{j}\varepsilon ^{kjn}L_{n}\\\end{aligned}}}$$ or in vector form, dividing by c $${\displaystyle \mathbf {N} '=\gamma \mathbf {N} -\left({\frac {\gamma -1}{\beta ^{2}}}\right){\boldsymbol {\beta }}\left({\boldsymbol {\beta }}\cdot \mathbf {N} \right)-{\frac {1}{c}}\gamma {\boldsymbol {\beta }}\times \mathbf {L} }$$ or reinstating β = v/c, $${\displaystyle \mathbf {N} '=\gamma \mathbf {N} -\left({\frac {\gamma -1}{v^{2}}}\right)\mathbf {v} \left(\mathbf {v} \cdot \mathbf {N} \right)-\gamma \mathbf {v} \times \mathbf {L} }$$ and $${\displaystyle {\begin{aligned}L'^{k\ell }&=\left({\Lambda ^{k}}_{i}{\Lambda ^{\ell }}_{0}-{\Lambda ^{k}}_{0}{\Lambda ^{\ell }}_{i}\right)cN^{i}+{\Lambda ^{k}}_{i}{\Lambda ^{\ell }}_{j}L^{ij}\\&=\gamma c\left(\beta ^{k}{\delta ^{\ell }}_{i}-\beta ^{\ell }{\delta ^{k}}_{i}\right)N^{i}+\leftL^{ij}\\&=\gamma c\left(\beta ^{k}N^{\ell }-\beta ^{\ell }N^{k}\right)+L^{k\ell }+{\frac {\gamma -1}{\beta ^{2}}}\beta ^{\ell }\beta _{j}L^{kj}+{\frac {\gamma -1}{\beta ^{2}}}\beta ^{k}\beta _{i}L^{i\ell }\\\end{aligned}}}$$ or converting to pseudovector form $${\displaystyle {\begin{aligned}\varepsilon ^{k\ell n}L'_{n}&=\gamma c\left(\beta ^{k}N^{\ell }-\beta ^{\ell }N^{k}\right)+\varepsilon ^{k\ell n}L_{n}+{\frac {\gamma -1}{\beta ^{2}}}\left(\beta ^{\ell }\beta _{j}\varepsilon ^{kjn}L_{n}-\beta ^{k}\beta _{i}\varepsilon ^{\ell in}L_{n}\right)\\\end{aligned}}}$$ in vector notation $${\displaystyle \mathbf {L} '=\gamma c{\boldsymbol {\beta }}\times \mathbf {N} +\mathbf {L} +{\frac {\gamma -1}{\beta ^{2}}}{\boldsymbol {\beta }}\times ({\boldsymbol {\beta }}\times \mathbf {L} )}$$ or reinstating β = v/c, $${\displaystyle \mathbf {L} '=\gamma \mathbf {v} \times \mathbf {N} +\mathbf {L} +{\frac {\gamma -1}{v^{2}}}\mathbf {v} \times \left(\mathbf {v} \times \mathbf {L} \right)}$$

Rigid body rotation

For a particle moving in a curve, the cross product of its angular velocity ω (a pseudovector) and position x give its tangential velocity $${\displaystyle \mathbf {u} ={\boldsymbol {\omega }}\times \mathbf {x} }$$

which cannot exceed a magnitude of c, since in SR the translational velocity of any massive object cannot exceed the speed of light c. Mathematically this constraint is 0 ≤ |u| < c, the vertical bars denote the magnitude of the vector. If the angle between ω and x is θ (assumed to be nonzero, otherwise u would be zero corresponding to no motion at all), then |u| = |ω| |x| sin θ and the angular velocity is restricted by $${\displaystyle 0\leq |{\boldsymbol {\omega }}|<{\frac {c}{|\mathbf {x} |\sin \theta }}}$$

The maximum angular velocity of any massive object therefore depends on the size of the object. For a given |x|, the minimum upper limit occurs when ω and x are perpendicular, so that θ = π/2 and sin θ = 1.

For a rotating rigid body rotating with an angular velocity ω, the u is tangential velocity at a point x inside the object. For every point in the object, there is a maximum angular velocity.

The angular velocity (pseudovector) is related to the angular momentum (pseudovector) through the moment of inertia tensor I $${\displaystyle \mathbf {L} =\mathbf {I} \cdot {\boldsymbol {\omega }}\quad \rightleftharpoons \quad L_{i}=I_{ij}\omega _{j}}$$ (the dot · denotes tensor contraction on one index). The relativistic angular momentum is also limited by the size of the object.

Spin in special relativity

Four-spin

A particle may have a "built-in" angular momentum independent of its motion, called spin and denoted s. It is a 3d pseudovector like orbital angular momentum L.

The spin has a corresponding spin magnetic moment, so if the particle is subject to interactions (like electromagnetic fields or spin-orbit coupling), the direction of the particle's spin vector will change, but its magnitude will be constant.

The extension to special relativity is straightforward. For some lab frame F, let F′ be the rest frame of the particle and suppose the particle moves with constant 3-velocity u. Then F′ is boosted with the same velocity and the Lorentz transformations apply as usual; it is more convenient to use β = u/c. As a four-vector in special relativity, the four-spin S generally takes the usual form of a four-vector with a timelike component s_t and spatial components s, in the lab frame $${\displaystyle \mathbf {S} \equiv \left(S^{0},S^{1},S^{2},S^{3}\right)=(s_{t},s_{x},s_{y},s_{z})}$$ although in the rest frame of the particle, it is defined so the timelike component is zero and the spatial components are those of particle's actual spin vector, in the notation here s′, so in the particle's frame $${\displaystyle \mathbf {S} '\equiv \left({S'}^{0},{S'}^{1},{S'}^{2},{S'}^{3}\right)=\left(0,s_{x}',s_{y}',s_{z}'\right)}$$

Equating norms leads to the invariant relation $${\displaystyle s_{t}^{2}-\mathbf {s} \cdot \mathbf {s} =-\mathbf {s} '\cdot \mathbf {s} '}$$ so if the magnitude of spin is given in the rest frame of the particle and lab frame of an observer, the magnitude of the timelike component s_t is given in the lab frame also.

The boosted components of the four spin relative to the lab frame are $${\displaystyle {\begin{aligned}{S'}^{0}&={\Lambda ^{0}}_{\alpha }S^{\alpha }={\Lambda ^{0}}_{0}S^{0}+{\Lambda ^{0}}_{i}S^{i}=\gamma \left(S^{0}-\beta _{i}S^{i}\right)\\&=\gamma \left({\frac {c}{c}}S^{0}-{\frac {u_{i}}{c}}S^{i}\right)={\frac {1}{c}}U_{0}S^{0}-{\frac {1}{c}}U_{i}S^{i}\\{S'}^{i}&={\Lambda ^{i}}_{\alpha }S^{\alpha }={\Lambda ^{i}}_{0}S^{0}+{\Lambda ^{i}}_{j}S^{j}\\&=-\gamma \beta ^{i}S^{0}+\leftS^{j}\\&=S^{i}+{\frac {\gamma ^{2}}{\gamma +1}}\beta _{i}\beta _{j}S^{j}-\gamma \beta ^{i}S^{0}\end{aligned}}}$$